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  • Let α ∈ (0,1) and β = loge(1 − α). Let   <img alt="P_n(x)=x+\frac{x^2}{2}+\frac{x^3}{3}+\cdots+\frac{x^n}{n}, x \in(0,1)" src="https://entrancecorner.oncodecog

Let α ∈ (0,1) and β = loge(1 − α). Let   P_n(x)=x+\frac{x^2}{2}+\frac{x^3}{3}+\cdots+\frac{x^n}{n}, x \in(0,1)
Then the integral   \int_0^\alpha \frac{t^{50}}{1-t} d t   is equal to

 

Option: 1

 β +  P_{50} (α)


Option: 2

P_{50} (α) − β


Option: 3

β − P_{50} (α)


Option: 4

−(β + P_{50} (α))


Answers (1)

best_answer

\begin{aligned} & \alpha \in(0,1), \beta=\log _e(1-\alpha) \\ & P_n(x)=x+\frac{x^2}{2}+\frac{x^3}{3}+\ldots \ldots+\frac{x^n}{n}, x \in(0,1) \\ & \int_0^\alpha \frac{t^{50}-1+1}{1-\mathrm{t}} \mathrm{dt} \\ & -\int_0^\alpha \frac{1-\mathrm{t}^{50}}{1-\mathrm{t}} \mathrm{dt}+\int_0^\alpha \frac{1}{1-\mathrm{t}} \mathrm{dt} \end{aligned}

\begin{aligned} & -\int_0^\alpha\left(1+\mathrm{t}+\mathrm{t}^2+\ldots \ldots \mathrm{t}^{49}\right) \mathrm{dt}-[\ln (1-\mathrm{t})]_0^\alpha \\ & -\left[\mathrm{t}+\frac{\mathrm{t}^2}{2}+\frac{\mathrm{t}^3}{3}+\ldots \ldots \frac{\mathrm{t}^{50}}{50}\right]_0^\alpha-\ln (1-\alpha) \\ & -\left[\alpha+\frac{\alpha^2}{2}+\frac{\alpha^3}{3}+\ldots \ldots+\frac{\alpha^{50}}{50}\right]-\ln (1-\alpha) \\ & -P_{50}(\alpha)-\ln (1-\alpha) \\ & -\left(\beta+P_{50}(\alpha)\right) \end{aligned}

 

Posted by

Ritika Jonwal

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